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1) I know that turing decidable means recursive language. But does is also means its decidable? So basically i want to know if REC imples decidability and RE implies undecidability or not. I got confused with word decidable in "turing decidable"

2) http://www.geeksforgeeks.org/np-completeness-set-1/ this links says that NP is those decision problems that can be solved by non-deterministic turing machines and P are those problems which can be solved by deterministic turing machines

So I thought that REC languages are accepted by DTM and RE by NDTM but it turned out that I was false according to this link https://gateoverflow.in/8111/gate2015-2_21 . So that means non-determinism have nothing to do with halting ? and that REC are accepted by DTM and NDTM both ? and also RE also accepted by DTM and NDTM both ?

3) Why we are concerned with DTM and NDTM in P and NP decision problems if we say that DTM and NDTM have same expressive power ? If they have same expressive power why can't we use DTM in NP decision problems?

Thanks for being patient and reading doubt.

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First of all you should first read these topics from any standard books and not from geeksforgeeks or GO for that matter. After you have read the basics then these sites will be useful.

  1. DTM and NDTM have same power -- meaning they both accept the same class of language. 
  2. P and NP as told in the question are decided respectively by deterministic and non-deterministic TMs in polynomial time. Yes, 'polynomial time' is the reason for the existence of these classes. Otherwise any problem in P or NP can be decided by a deterministic TM.
  3. Non-deterministic TM cannot decide any problem which is undecidable.
  4. Both deterministic and non-deterministic TMs can accept the strings of any r.e. language.
  5. The language of an undecidable language can still be r.e. (semi-decidable) -- halting problem is an example.
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